If 20! × 20!/20^n is an integer, what is the largest possible value of

Question

If 20! × 20!/20^n is an integer, what is the largest possible value of n?

A. 20 
B. 16 
C. 8 
D. 4 
E. 1

OptimusPrepJanielle · Accepted Answer

20! × 20!/20^n is an integer.
Or 20! × 20!/(2^2n*5^n)

This means we need to find the power of 5 in the numerator. We can safely say that the number of powers of 5 will be lower than the number of powers of 4 in the numerator.

Largest power of 5 in 20! =   +   = 4
Since there are 2 20!'s, power of 4 in the numerator = 2*4 = 8

Hence the largest value of b for which 20! × 20!/20^n is integer = 8

Correct Option : C

nipunjain14 · Answer

in 20! following can be factorized as 20 :
1*20, 2*10, 4*5

also another factor from 5*3*8 can be divided by 20 
so n = 8

Ans: C

stonecold · Answer

This is an Excellent Question.
Here number of 5's plays a limiting role.
As the number of 4's would be sufficient. (ample number of 2's)
number of 5's in 20! => 20/5+20/25 => 4
therefore 20!*20! would have 8 fives and ample number of 4's 
so since a 20 is formed by a 5 and a +> Maximum number of 20's => 8
N can be almost 8

Hence C

destinyawaits · Answer

could you explain how 20/5 +20/25 equals four?

Bunuel · Answer

You should take only the quotient of the division, that is 20/5 = 4 and 20/25 = 0.

dieplengoc · Answer

As 20=2^2 * 5
(20!)^2 = (1*2*...*20)^2 = (2^10*5^4*...)^2 = (20^4*...)^2 = 20^8*...
Hence C.

MathRevolution · Answer

20^n  =  (2 * 2 * 5)^n

=> 20 =  (2^2)^n * 5^n

So, we need a number of 5's in 20! to find the value of n.

=>  \frac{20}{5}  +  \frac{(20) }{ 5^2 }

=> 4 + 0

=> So there will be total 4 + 4 = 8 (5's) for

Therefore n = 8

Answer C

datbanhbao · Answer

I dont know this will this solution be correct or not? But I hope I can share to you because I got wrong with this quiz

20! x 20!/20^n = Integer >> this one has to be divisible >> we have: (20!)^2/20^n = integer
20! will have 4 pairs that can divide 20: 1-20; 2-10; 4-5; 15-8 >> (20!)^2 will have 8 pairs >> n = 8 >> C

Crytiocanalyst · Answer

20! will have 4 20s
since we can only make 20s 
with 5 and only 4 multiples 5, 10 ,15,20 exists 
THey can be made 20 with the it's respective multiple of 2 
and 2 20! exists therefore the total number becomes 8
Therefore IMO C

AJ0784 · Answer

Think in this way to understand this question:
For 625/5^n to be an integer the max value of n can be 4 since 625=5^4
Similarly for 20!*20!/20^4 to be an integer we have to find out the max value of the power of 5 in 20!*20!.
We have 4 fives in 20! Therefore 20!*20!has 8 fives.Hence the maximum value of n can be 8.

Posted from my mobile device

BraydinWalters · Answer

Here's how I solved...

we have:

20! x 20! 
 --- 20

Prime factorization of 20 = (2 x 2 x 5)

Now let's see how many 2s and 5s are in 20!:

20/2 = 10
10/ 2 = 5
5/2 = 2 (only count number of times it divides into it without remainders)
2/2 = 1
1/2 = 0

10 + 5 + 2 + 1 = 18
There are 18 2s in 20! and 36 in 20! x 20!

Now for how many 5s are in 20!:

20/5 = 4
4/5 = 0

There are 4 5s in 20!

Remember we have 20! x 20! so 4 + 4 = 8 5s in total.

With one 5 in 20, we can raise it to the power of 8 (20^8) to still get an integer when dividing 20! + 20!

8 is your answer

DonBosco7 · Answer

What applies to this question is something called as  Finding the power of a prime number in a factorial 
=> For example:- If we have to find the power of a prime number 'p' in n! then the power could be found using
 + + +............

we are given 20!*20!/20^n

Now the denominator as everybody said it will be (5*4)^n, But here mind you 4 is not prime, but 4 = 2^2 and 2 is a prime.
So when we solve using the method describe above. 20/5 = 4
Now there will be as many 4s as 5s, so the power of (5*4)^n = (5*4)^4 = 5^4 * 4^4
but as we have said above 4 is not prime but 2 is, hence 4^4 becomes 2^2^4 = 2^8
Hence your highest power is 8, since 5 only gave a power of 4 and 2 gave a power of 8, hence C.

If 20! × 20!/20^n is an integer, what is the largest possible value of

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If 20! × 20!/20^n is an integer, what is the largest possible value of

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